If `n(A_(i))=i+1and A_(1)subA_(2)sub... subA_(99),then n(underset(i-1)overset(99)UA_(i))=`

Options

A. 21
B. 7
C. 100
D. 14

Answer

To solve the given problem, we need to find the number of elements in the union of the sets \( A_1, A_2, \ldots, A_{99} \), where the number of elements in each set \( A_i \) is given by \( n(A_i) = i + 1 \) and the sets are nested such that \( A_1 \subset A_2 \subset \ldots \subset A_{99} \).

### Step-by-Step Solution:

1. **Understand the Number of Elements in Each Set**:
- We are given that \( n(A_i) = i + 1 \).
- This means:
- \( n(A_1) = 1 + 1 = 2 \)
- \( n(A_2) = 2 + 1 = 3 \)
- \( n(A_3) = 3 + 1 = 4 \)
- ...
- \( n(A_{99}) = 99 + 1 = 100 \)

2. **Identify the Relationship Between the Sets**:
- The problem states that \( A_1 \subset A_2 \subset \ldots \subset A_{99} \).
- This implies that all elements of \( A_1 \) are in \( A_2 \), all elements of \( A_2 \) are in \( A_3 \), and so on, up to \( A_{99} \).
- Therefore, \( A_{99} \) contains all elements from \( A_1 \) to \( A_{98} \).

3. **Calculate the Union of the Sets**:
- The union of all sets from \( A_1 \) to \( A_{99} \) can be expressed as:
\[ \bigcup_{i=1}^{99} A_i = A_{99} \]
- This is because \( A_{99} \) contains all the elements of the previous sets.

4. **Find the Number of Elements in the Union**:
- Since \( \bigcup_{i=1}^{99} A_i = A_{99} \), we need to find \( n(A_{99}) \).
- From our earlier calculation, we know that:
\[ n(A_{99}) = 100 \]

5. **Conclusion**:
- Therefore, the number of elements in the union of the sets \( A_1, A_2, \ldots, A_{99} \) is:
\[ n\left(\bigcup_{i=1}^{99} A_i\right) = n(A_{99}) = 100 \]

### Final Answer:
The number of elements in the union \( \bigcup_{i=1}^{99} A_i \) is \( 100 \).

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